Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $k = \dfrac{z^2 + 4z - 60}{z^2 + z} \times \dfrac{7z + 7}{-3z - 30} $
Solution: First factor the quadratic. $k = \dfrac{(z + 10)(z - 6)}{z^2 + z} \times \dfrac{7z + 7}{-3z - 30} $ Then factor out any other terms. $k = \dfrac{(z + 10)(z - 6)}{z(z + 1)} \times \dfrac{7(z + 1)}{-3(z + 10)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ (z + 10)(z - 6) \times 7(z + 1) } { z(z + 1) \times -3(z + 10) } $ $k = \dfrac{ 7(z + 10)(z - 6)(z + 1)}{ -3z(z + 1)(z + 10)} $ Notice that $(z + 1)$ and $(z + 10)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ 7\cancel{(z + 10)}(z - 6)(z + 1)}{ -3z(z + 1)\cancel{(z + 10)}} $ We are dividing by $z + 10$ , so $z + 10 \neq 0$ Therefore, $z \neq -10$ $k = \dfrac{ 7\cancel{(z + 10)}(z - 6)\cancel{(z + 1)}}{ -3z\cancel{(z + 1)}\cancel{(z + 10)}} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $k = \dfrac{7(z - 6)}{-3z} $ $k = \dfrac{-7(z - 6)}{3z} ; \space z \neq -10 ; \space z \neq -1 $